Question

How many real solutions if any, does 2x^2-3x+8=0​

Answer 1

Answer:

The quadratic equation $2\, x^{2} - 3\, x + 8 = 0$ has no real solution.

Step-by-step explanation:

Rewrite the quadratic equation $2\, x^{2} - 3\, x + 8 = 0$ in the standard form $a\, x^{2} + b\, x + c = 0$:

$2\, x^{2} + (-3)\, x + 8 = 0$, for which:

• $a = 2$.
• $b = (-3)$.
• $c = 8$.

The quadratic discriminant of $a\, x^{2} + b\, x + c = 0$ is $(b^{2} - 4\, a\, c)$.  The quadratic discriminant of $2\, x^{2} + (-3)\, x + 8 = 0$ would be:

$\begin{aligned}& b^{2} - 4\, a\, c \\ =\; & (-3)^{2} - 4 \times 2 \times 8 \\ =\; & (-55)\end{aligned}$.

Since the quadratic discriminant of this equation is negative, this quadratic equation has no real solution.